#inte^x[((sin^-1x)sqrt(1-x^2)+1)/sqrt(1-x^2)]dx# ?

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Answer 1 · 2016-07-28T16:03:41

#e^x(sin^-1x)+C#

We have:

#inte^x[((sin^-1x)sqrt(1-x^2)+1)/sqrt(1-x^2)]dx#

Splitting up the fraction, this becomes:

#=inte^x[sin^-1x+1/sqrt(1-x^2)]dx#

Split up the integrals:

#=inte^x(sin^-1x)dx+inte^x/sqrt(1-x^2)dx#

Attempt to integrate #inte^x(sin^-1x)dx# via integration by parts, which takes the form #intuv=uv-intvdu#.

Let #u=sin^-1x#, so #du=1/sqrt(1-x^2)dx#, and #dv=e^xdx#, so #v=e^x#.

Thus we see that the original integral equals:

#=[e^x(sin^-1x)-inte^x/sqrt(1-x^2)dx]+inte^x/sqrt(1-x^2)dx#

The integral #inte^x/sqrt(1-x^2)dx# will cancel with itself, leaving just

#=e^x(sin^-1x)#

Add the constant of integration:

#=e^x(sin^-1x)+C#

Answer 2 · 2016-07-28T16:07:42

#e^x(sin^-1x)+C#

There is a pattern surrounding integrals involving #e^x#, in cases like this problem (and all the others surrounding it on the page).

Since:

#d/dxe^xf(x)=e^xf(x)+e^xf'(x)=e^x[f(x)+f'(x)]#

Then:

#inte^x[f(x)+f'(x)]dx=e^xf(x)+C#

This is the case for the given problem, which simplifies to be:

#inte^x[sin^-1x+1/sqrt(1-x^2)]dx#

If #f(x)=sin^-1x#, then #f'(x)=1/sqrt(1-x^2)#.

Thus the integral equals

#=e^xf(x)+C=e^x(sin^-1x)+C#